Answer:
![PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]](https://tex.z-dn.net/?f=PV_%7Bm%7D%20%3D%20RT%5B1%20%2B%20%28b-%5Cfrac%7Ba%7D%7BRT%7D%29%5Cfrac%7B1%7D%7BV_%7Bm%7D%20%7D%20%2B%20%5Cfrac%7Bb%5E%7B2%7D%20%7D%7BV%5E%7B2%7D%20_%7Bm%7D%20%7D%20%2B%20...%5D)
B = b -a/RT
C = b^2
a = 1.263 atm*L^2/mol^2
b = 0.03464 L/mol
Explanation:
In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:
Using the van deer Waals equation of state:

With further simplification, we have:
![P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]](https://tex.z-dn.net/?f=P%20%3D%20RT%5B%5Cfrac%7B1%7D%7BV_%7Bm%7D-b%20%7D%20-%20%5Cfrac%7Ba%7D%7BRTV_%7Bm%7D%20%5E%7B2%7D%20%7D%5D)
Then, we have:
![P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BRT%7D%7BV_%7Bm%7D%20%7D%20%5B%5Cfrac%7B1%7D%7B1-%5Cfrac%7Bb%7D%7BV_%7Bm%7D%20%7D%20%7D%20-%20%5Cfrac%7Ba%7D%7BRTV_%7Bm%7D%20%7D%5D)
Therefore,
![PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]](https://tex.z-dn.net/?f=PV_%7Bm%7D%20%3D%20RT%5B%281-%5Cfrac%7Bb%7D%7BV_%7Bm%7D%20%7D%29%20%5E%7B-1%7D%20-%20%5Cfrac%7Ba%7D%7BRTV_%7Bm%7D%20%7D%5D)
Using the expansion:

Therefore,
![PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]](https://tex.z-dn.net/?f=PV_%7Bm%7D%20%3D%20RT%5B1%2B%5Cfrac%7Bb%7D%7BV_%7Bm%7D%20%7D%2B%5Cfrac%7Bb%5E%7B2%7D%20%7D%7BV_%7Bm%7D%20%5E%7B2%7D%20%7D%20%2B%20...%20-%5Cfrac%7Ba%7D%7BRTV_%7Bm%7D%20%7D%5D)
Thus:
equation (1)
Using the virial equation of state:
![P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]](https://tex.z-dn.net/?f=P%20%3D%20RT%5B%5Cfrac%7B1%7D%7BV_%7Bm%7D%20%7D%2B%20%5Cfrac%7BB%7D%7BV_%7Bm%7D%20%5E%7B2%7D%7D%2B%5Cfrac%7BC%7D%7BV_%7Bm%7D%20%5E%7B3%7D%20%7D%2B%20...%5D)
Thus:
equation (2)
Comparing equations (1) and (2), we have:
B = b -a/RT
C = b^2
Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.
[/tex] = 0.03464 L/mol
a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2
Answer:
Mass = 0.697 g
Explanation:
Given data:
Volume of hydrogen = 1.36 L
Mass of ammonia produced = ?
Temperature = standard = 273.15 K
Pressure = standard = 1 atm
Solution:
Chemical equation:
3H₂ + N₂ → 2NH₃
First of all we will calculate the number of moles of hydrogen:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
1atm ×1.36 L = n × 0.0821 atm.L/mol.K × 273.15 K
1.36 atm.L = n × 22.43 atm.L/mol
n = 1.36 atm.L / 22.43 atm.L/mol
n = 0.061 mol
Now we will compare the moles of hydrogen and ammonia:
H₂ : NH₃
3 : 2
0.061 : 2/3×0.061 = 0.041
Mass of ammonia:
Mass = number of moles × molar mass
Mass = 0.041 mol × 17 g/mol
Mass = 0.697 g
The % yield of Ca(OH)₂ : 62.98%
<h3>Further eplanation
</h3>
Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated
General formula:
Percent yield = (Actual yield / theoretical yield )x 100%
An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients
Reaction
CaO + H₂O ⇒ Ca(OH)₂
mass CaO= 4.2 g
mol CaO(MW=56,0774 g/mol) :

mol Ca(OH)₂ based on mol CaO
mol ratio CaO : Ca(OH)₂,= 1 : 1, so mol Ca(OH)₂ = 0.075
mass Ca(OH)₂(MW=74,093 g/mol) ⇒ theoretical

% yield :

The correct answer is Solute
Explanation:
In chemistry, a solution refers to a homogenous mixture of two substances that occurs through dissolution, this means once they are mixed the substances form a uniform new substance and cannot be easily separated. Additionally, in chemistry, the substances involved in a solution are either classified as solutes if they are the substances that dissolve to form a solution or as solvents in the case of substances in which the solute dissolves in. For example, if you mix salt and water, the salt acts as the solute while the water is the solvent. Thus, the component which dissolves in a solution is called the solute.